Suppose that a 15-mL sample of a solution is to be tested for I- ion by addition of 1 drop (0.20 mL) of 0.30 M Pb(NO3)2. What is the minimum number of grams of I- that must be present for PbI2(s) to form? (Ksp = 7.9 10-9)
PbI2 ==> Pb^2+ + 2I^-
Ksp = (Pb^2+)(I^-)^2
0.2mL is a big drop but the concn Pb^2+ in the 15 mL sample will be 0.3M x 0.2/15 = ?
Substitute Ksp, ? for Pb^2+, and solve for (I^-) in moles/L.
Convert to moles in 15 mL, then to grams. g = mols x molar mass.
To find the minimum number of grams of I- that must be present for PbI2(s) to form, we need to determine the limiting reactant and calculate the amount of I- needed using stoichiometry.
Step 1: Write the balanced equation for the reaction between Pb(NO3)2 and I-:
Pb(NO3)2 + 2I- -> PbI2(s) + 2NO3-
Step 2: Calculate the amount of Pb(NO3)2 added in moles:
Volume of Pb(NO3)2 added = 0.20 mL = 0.20/1000 L = 0.0002 L
Concentration of Pb(NO3)2 = 0.30 M
moles of Pb(NO3)2 = concentration x volume = 0.30 x 0.0002 = 0.00006 moles
Step 3: Determine the stoichiometry between Pb(NO3)2 and I-:
From the balanced equation, we see that 1 mole of Pb(NO3)2 reacts with 2 moles of I-. Therefore, the moles of I- needed will be twice the moles of Pb(NO3)2.
Step 4: Calculate the moles of I- needed:
moles of I- needed = 2 x moles of Pb(NO3)2 = 2 x 0.00006 = 0.00012 moles
Step 5: Convert moles of I- to grams:
The molar mass of I- is 127 g/mol.
grams of I- = moles of I- x molar mass of I-
grams of I- = 0.00012 moles x 127 g/mol = 0.01524 grams
Therefore, the minimum number of grams of I- that must be present for PbI2(s) to form is 0.01524 grams (or approximately 15.24 mg).
To find the minimum number of grams of I- that must be present for PbI2(s) to form, we need to calculate the maximum amount of PbI2 that can be formed based on the given Ksp value.
Step 1: Convert the volume of the Pb(NO3)2 solution added to moles of Pb(NO3)2.
moles of Pb(NO3)2 = volume (L) x concentration (mol/L)
moles of Pb(NO3)2 = 0.20 mL x (1 L / 1000 mL) x 0.30 mol/L
moles of Pb(NO3)2 = 0.00006 mol
Step 2: Since the stoichiometry of the reaction is 1:2 (1 Pb(NO3)2 to 2 I-), the number of moles of I- ions that can react is twice the moles of Pb(NO3)2 added.
moles of I- = 2 x moles of Pb(NO3)2
moles of I- = 2 x 0.00006 mol
moles of I- = 0.00012 mol
Step 3: Calculate the minimum mass of I- ions that must be present.
mass of I- = moles of I- x molar mass of I-
mass of I- = 0.00012 mol x 127.0 g/mol
mass of I- = 0.01524 g
Therefore, the minimum number of grams of I- that must be present for PbI2(s) to form is 0.01524 grams.